Recently I handed in a problem set containing the following question, but neither myself nor my classmates managed to find a satisfying solution. We were quite certain that a fruitful approach was to use Strassen's theorem (see this post for a statement of it), but the details were elusive.

Suppose that $X$ and $Y$ are random variables on some probability space such that for all $\delta, \varepsilon>0$, $$ \mathbb{P}(X\in A) \le \mathbb{P}(Y \in A^\delta) + \varepsilon $$ for each Borel set $A \subseteq \mathbb{R}$. If $U$ is a random variable that is independent of $(X,Y)$ and is uniformly distributed on $[0,1]$, prove that there exists a measurable function $f:\mathbb{R}\times [0,1] \to \mathbb{R}$ such that $$ \mathbb{P}(|X - f(X,U)| > \delta) \le \varepsilon $$ and that $f(X,U)$ has the same distribution as $Y$.

Letting $d$ denote the Euclidean metric, we noticed that by Strassen's theorem on $\mathbb{P}\circ X^{-1}$ and $\mathbb{P}\circ Y^{-1}$ we could obtain, for each $\varepsilon' > 0$, nonnegative measures $\eta$ and $\gamma$ on $\mathbb{R}^2$ such that $\eta \{(x,y) \in \mathbb{R}^2: d(x,y) > \delta + \varepsilon' \} = 0$ and $\gamma (\mathbb{R}^2) \le \varepsilon + \varepsilon'$. In particular, this meant that for $\mu = \eta + \gamma$ we had $$ \mu( d(x,y) > \delta + \varepsilon') \le \mu(\mathbb{R}^2) \le \varepsilon + \varepsilon' $$

Based on this, our general approach to the question was as follows:

  1. For each $n\ge 1$, apply Strassen's theorem to obtain a sequence of measures $(\mu_n) = (\eta_n + \gamma_n)$ such that each $\mu_n$ has marginals $\mathbb{P}\circ X^{-1}$ and $\mathbb{P}\circ Y^{-1}$ and $$ \mu_n( d(x,y) > \delta + 1/n) \le \eta_n( d(x,y) > \delta + 1/n) + \gamma_n( d(x,y) > \delta + 1/n) \le \varepsilon + 1/n $$

  2. Ulam's theorem says that $\mathbb{P}\circ X^{-1}$ and $\mathbb{P}\circ Y^{-1}$ are tight, so there's a compact set $K\subset \mathbb{R}$ for which $\mu_n(K\times K) \ge 1-2\delta$. Hence $(\mu_n)$ is uniformly tight and the Helly selection theorem says that the limit measure $\mu$ has marginals $\mathbb{P}\circ X^{-1}$ and $\mathbb{P}\circ Y^{-1}$.

  3. $\{d(x,y) > \delta + 1/n\}$ is open in $\mathbb{R}^2$, so with the portmanteau theorem, we may conclude that $\mu\{d(x,y) > \delta \} \le \varepsilon$.

  4. All that seems to remain is to produce a function $f$ for which $$ \mu(d(x,y) > \delta) = \mathbb{P}(d(X,f(X,U) > \delta) $$ and check that it has the required properties.

Step 4 is where we got stuck. Any ideas or solutions would be appreciated.

  • $\begingroup$ In Kalleneberg there is a statement that every conditional law can be represented as a function $f(\cdot,U)$ where $U$ is e.g. uniform. $\endgroup$ – Ilya Dec 7 '14 at 22:45
  • $\begingroup$ I assume you're referring to Theorem 6.10, but is it then obvious that this function satisfies $\mu(d(x,y)> \delta) = \mathbb{P}(d(X,f(X,U)>\delta )$? $\endgroup$ – sourisse Dec 8 '14 at 15:52
  • $\begingroup$ Well, I'd really expect that the problem has to be solved with $\mu = \Bbb P$, otherwise there is no much connection between them besides having the same marginals. $\endgroup$ – Ilya Dec 8 '14 at 16:16

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